TapeEquilibrium

A non-empty zero-indexed array A consisting of N integers is given. Array A represents numbers on a tape. Any integer P, such that 0 < P < N, splits this tape into two non−empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1]. The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])| In other words, it is the absolute difference between the sum of the first part and the sum of the second part. For example, consider array A such that: A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3 We can split this tape in four places: P = 1, difference = |3 − 10| = 7 P = 2, difference = |4 − 9| = 5 P = 3, difference = |6 − 7| = 1 P = 4, difference = |10 − 3| = 7 Write a function: class Solution { public int solution(int[] A); } that, given a non-empty zero-indexed array A of N integers, returns the minimal difference that can be achieved. For example, given: A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3 the function should return 1, as explained above.